Let $g(x)=\begin{cases} \dfrac{1}{x}&\text{for }x\leq -2 \\\\ -\dfrac{\cos(x+2)}{2}&\text{for }x> -2 \end{cases}$ Is $g$ continuous at $x=-2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
For $g$ to be continuous at $x=-2$, we need $\lim_{x\to -2}g(x)$ and $g(-2)$ to exist and be equal. Since $-2\leq -2$, the rule that applies to $x=-2$ is $\dfrac{1}{x}$. So $g(-2)=\dfrac{1}{-2}=-\dfrac{1}{2}$. Now let's analyze $\lim_{x\to -2}g(x)$. Finding $\lim_{x\to -2^{ +}}g(x)$ For $x$ -values larger than $-2$, the appropriate rule for $g(x)$ is $-\dfrac{\cos(x+2)}{2}$. Since $-\dfrac{\cos(x+2)}{2}$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to -2^{ +}}g(x) \\\\ &=\lim_{x\to -2^{ +}}-\dfrac{\cos(x+2)}{2} \gray{-\dfrac{\cos(x+2)}{2}\text{ is the rule for }x>-2} \\\\ &=-\dfrac{\cos(-2+2)}{2} \gray{-\dfrac{\cos(x+2)}{2}\text{ is continuous at }x=-2} \\\\ &=-\dfrac{1}{2} \end{aligned}$ Finding $\lim_{x\to -2^{ -}}g(x)$ For $x$ -values smaller than $-2$, the appropriate rule for $g(x)$ is $\dfrac{1}{x}$. Since $\dfrac{1}{x}$ is continuous for $x\leq -2$, any limit in this interval is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to -2^{ -}}g(x) \\\\ &=\lim_{x\to -2^{ -}}\dfrac{1}{x} \gray{\dfrac{1}{x}\text{ is the rule for }x<-2} \\\\ &=\dfrac{1}{-2} \gray{\dfrac{1}{x}\text{ is continuous at }x=-2} \\\\ &=-\dfrac{1}{2} \end{aligned}$ Conclusion We found that: $\lim_{x\to -2^{ +}}g(x)=\lim_{x\to -2^{ -}}g(x)=g(-2)=-\dfrac{1}{2}$ Since the one-sided limits are both equal to $g(-2)$, we can determine that the two-sided limit $\lim_{x\to -2}g(x)$ is also equal to $g(-2)$, and $g$ is continuous at $x=-2$.